googologytestingfandomcom-20200215-history
User blog:Edwin Shade 2/Formula For Prime Numbers
In this user page, I presented a rough sketch of the Ж-OCF. In this blog post however, I will be eschewing the typical course of examining \(Ж(\alpha)\) when \(\alpha\) is transfinite, and actually restrict myself to talking about values of \(\alpha\in\mathbb{N}\) in the expression \(Ж(\alpha)\). This is to expound upon some interesting behavior I discovered when evaluating successive values of \(Ж(\alpha)\) for finite natural numbers. Firstly, please go back to the userpage I linked if you are unclear about how the \(Ж(\alpha)\) function works. It is not even a fully fleshed out OCF at this point so it is only neccesary to get the very basics. Anyhow, \(Ж(0)\) is just 1, because since \(P_0=\{0,\Omega\}\) and \(Ж(\alpha)\) denotes the first ordinal not in \(P_{\alpha}\), clearly the first ordinal not in \(P_0\) is 1. But from here on out the behavior gets interesting. You see, for all \(Ж(\alpha)\) where \(\alpha<\omega\) (or more formally, for all \(Ж(\alpha)\) beneath \(Ж(\Omega)\)), the result is a finite number. So we can disregard the transfinite numbers and really treat this as a matter of how to generate as many consecutive numbers from 0 to some number n given the set of \(P_{m}\), using only addition, multiplication, and exponentiation. Accordingly, \(Ж(m)\) will be \(n+1\), or the first natural inaccessible by this means. I will demonstrate this by plotting out the derivation of Ж numbers 0 through 8. *\(Ж(0)=1\) given *\(Ж(1)=2\) is given, \(P_1\) contains \(\Omega^0 = 1\) though *\(Ж(2)=3\) and 1 are given, 1 + 1 = 2 *\(Ж(3)=5\) though 2 are given, 3 = 1 + 2, 4 = 2^2 *\(Ж(4)=10\) through 4 are given, 5 = 1 + 4, 6 = 2 + 4, 7 = 3 + 4, 8 = 4 + 4, 9 = 3^2 *\(Ж(5)=19\) through 9 are given, 10 = 1 + 9, 11 = 2 + 9, 12 = 3 + 9, ..., 18 = 9 + 9 *\(Ж(6)=37\) through 18 are given, 19 = 1 + 18, 20 = 2 + 18, ..., 36 = 18 + 18 *\(Ж(7)=73\) through 36 are given, 37 = 1 + 36, 38 = 2 + 36, ..., 72 = 36 + 36 *\(Ж(8)=146\) through 72 are given, 73 = 1 + 72, 74 = 2 + 72, ..., 144 = 72 + 72, 145 = 5 x 29 So what can we conclude? Well first of all, it's clear that \(Ж(n)\) grows at least as quickly as \(2^n\), because of how we can use addition to always double \(P_n\) when making \(P_{n+1}\). But second of all, for every case that \(Ж(n+1)\neq Ж(n)\cdot 2\), \(Ж(n+1)\) tends to be a prime number! Eliminating multiples of two and the trival \(Ж(0) = 1\) therefore, we have the sequence 2, 3, 5, 19, 37, 73, ... Of course, in mathematics there are no coincidences. So let's think about why these prime numbers appear. Well, we already know that \(Ж(n+1)\) will be equal to \(2^n-1\) or greater. This means that \(Ж(n)>\sqrt{Ж(n+1)}\) most of the time, because in order for \(Ж(n)<\sqrt{Ж(n+1)}\) then \(Ж(n+1)>Ж(n)^2\). This is relevant because if you divide a number n by all \(m<\sqrt{n}\) and return no quotients then n is prime. Which means that in order for \(Ж(n+1)\) to not be prime, it either has to be \(Ж(n)\cdot 2\), or, \(Ж(n+1)\) must be at least \(Ж(n)^2\). For successive \(Ж(n)\), the ratio of \(\frac{Ж(n)^2}{Ж(n)\cdot 2}\) increases, decreasing the chance of all the numbers between \(Ж(n)\cdot 2\) and \(Ж(n)^2\) from being all attainable by the set of all numbers \(Ж(n-1)-1\) or less. This means \(Ж(n)\) tends to be prime, on the fault of probability. So this isn't really a sure-fire formula for prime numbers, but it definitely does generate primes most of the time. Of course a lot of factorization is needed to evaluate high values of \(Ж(n)\), so it's no a free lunch, but I just found this interesting. Category:Blog posts